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Lorentz transforms

  1. Aug 12, 2007 #1
    Zwiebach page 36

    a)Give the Lorentz transformations for the components a_mu of a vector under a boost along the x^1 axis.

    [tex]a'_0 = -a_0 \gamma + a_1 \gamma \beta [/tex]
    [tex]a'_1^ = -a_0 \gamma \beta + a_1 \gamma[/tex]
    and 2 and 3 are the same.

    b) Show that the objects

    [tex] \partial/\partial x^\mu[/tex] transform under a boost along the x^1 axis in the same way as the [tex]a_\mu[/tex] in (a) do

    Note the upper and lower indices.

    Firstly is my answer to (a) good? Second, I am not sure how to transform a partial derivative in (b).
     
  2. jcsd
  3. Aug 13, 2007 #2

    George Jones

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    I get a different result. Careful with th signs.

    Use the multivariable chain rule.
     
  4. Aug 13, 2007 #3
    Is this better for part (a)
    [tex]a'_0 = a_0 \gamma + a_1 \gamma \beta [/tex]
    [tex]a'_1 = a_0 \gamma \beta + a_1 \gamma[/tex]

    ?
     
  5. Aug 14, 2007 #4

    George Jones

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    Yes, this is what I get.
     
  6. Aug 17, 2007 #5
    I get, for the example of mu = 0:

    [tex] \frac{\partial}{\partial x'^0} = \frac{\partial}{\partial x'^0} \frac{\partial x' ^ 0}{\partial x^0} + \frac{\partial}{\partial x'^0} \frac{\partial x' ^ 0}{\partial x^1} [/tex]

    How do I get partials with respect to non-primed coordinates with the multivariable chain rule?
     
  7. Aug 17, 2007 #6

    George Jones

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    Do you know a formula that expresses [itex]x'^0[/itex] in terms of [itex]x^0[/itex] and [itex]x^1?[/itex] If you do, then just differentiate.
     
  8. Aug 18, 2007 #7
    Like this:

    [tex] \frac{\partial x'^0}{\partial x^0} = \gamma [/tex]

    [tex] \frac{\partial x'^0}{\partial x^1} = -\gamma \beta [/tex]

    In order to show that the objects

    [tex] \partial/\partial x^\mu[/tex]

    transform as suggested, I need to express the partial operators with respect to the primed coordinates in terms of the partial operators with respect to the nonprimed coordinates, right?

    What I showed above does not even have an operator really.
     
  9. Aug 20, 2007 #8

    George Jones

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    Substitute these into a corrected version of the equation that you gave in post #5. I just noticed that this equation is not right.
     
  10. Aug 23, 2007 #9
    I honestly do not see why my multivariable chain rule equation is wrong.
     
  11. Aug 23, 2007 #10

    George Jones

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    In post #5, you wrote

    "Cancellation" (I don't really mean cancellation) should turn each term on the right into the term on the left. After "cancellation' your expression is

    [tex] \frac{\partial}{\partial x'^0} = \frac{\partial}{\partial x^0} + \frac{\partial }{\partial x^1}. [/tex]

    Notice that each term on the right looks different than the term on the left.

    Take a close look at the chain rule in a text.
     
  12. Aug 24, 2007 #11
    I see. I am good with part (b). Now, for part (c), he means the expressions

    p = -i h-bar del

    and E = i h-bar d/dt

    right?

    So the p_mu is the momentum four-vector? How can you have a momentum four-vector when one component of that vector is basically time?
     
  13. Aug 28, 2007 #12

    George Jones

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    With lower indices. Forget about quantum theory for a bit. In an inertial frame, what is p_0?
     
  14. Aug 28, 2007 #13
    [tex] p_0 = -E/c [/tex]

    So I need to show that this is also equal to [tex] \hbar/i \frac{\partial}{\partial x^{\mu}} [/tex]? Should I use those equation in my previous post? Plugging in for E does not seem to work.
     
    Last edited: Aug 28, 2007
  15. Aug 29, 2007 #14

    George Jones

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    You need to show that it is equal to

    [tex] \hbar/i \frac{\partial}{\partial x^0} [/tex]

    Yes.
     
  16. Aug 29, 2007 #15
    Got it. Done with 2.3!
     
  17. Aug 30, 2007 #16

    George Jones

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    Great!

    Note that Zwiebach uses "physicist's math", i.e., he treats each [itex]\partial / \partial x^\mu[/itex] as a component of a covector. In modern (not so modern now) differential geometry, each [itex]\partial / \partial x^\mu[/itex] is a tangent vector, not a component.
     
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